What is the time complexity of Stone Game?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Stone Game?

This problem uses Array, Math, Dynamic Programming, Game Theory. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Stone Game?

Insight 1: Recognizing that the sum of stones is odd implies that the sum of stones at even indices cannot be equal to the sum of stones at odd indices. Insight 2: Understanding that Alice can always guarantee collecting all the stones at either the odd or even indices.

Which companies ask LeetCode 877. Stone Game in interviews?

This problem is frequently asked at Cisco.

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Stone Game - LeetCode 877 Solution

Stone Game - Complete Solution Guide

Stone Game is LeetCode problem 877, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Alice and Bob play a game with piles of stones. There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles[i] . The objective of the game is to end with the most stones. The total number of stones across all the piles is odd , so there are no ties. Alice and Bob take turns, with Alice starting first . Each turn, a player takes the entire pile of stones either from the beginning or from the end of the row. This continues until there are no mor

Detailed Explanation

The Stone Game involves two players, Alice and Bob, competing to collect the most stones from an even number of piles arranged in a row. Alice goes first, and on each turn, a player can take either the leftmost or rightmost pile. The game continues until all piles are taken. The player with the higher total number of stones wins. The problem asks whether Alice can win assuming both players play optimally, given the constraint that the total number of stones is odd.

Solution Approach

The solution leverages the fact that Alice can always win. Because the number of piles is even and the total number of stones is odd, Alice can compute the sum of stones at even indices and the sum of stones at odd indices. Whichever sum is larger, Alice can force the game to allow her to pick up all piles at those indices. Therefore, the solution simply returns `true`.

Step-by-Step Algorithm

Step 1: Analyze the problem constraints. The key constraint is that the total number of stones is odd, and there is an even number of piles.
Step 2: Realize that, given the above constraints, the sum of values at even indices and the sum of values at odd indices will never be equal.
Step 3: Understand that Alice always has the ability to pick either the first pile or last pile at each turn. This allows her to prioritize either even or odd indices.
Step 4: Conclude that Alice will always win, since she will pick the piles at either even or odd indices which give her a higher sum.
Step 5: Return `true`.

Key Insights

Insight 1: Recognizing that the sum of stones is odd implies that the sum of stones at even indices cannot be equal to the sum of stones at odd indices.
Insight 2: Understanding that Alice can always guarantee collecting all the stones at either the odd or even indices.
Insight 3: The realization that Alice always wins because she can choose which set of indices to prioritize (odd or even) based on which has the larger sum, ensuring her total is greater than Bob's.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Dynamic Programming, Game Theory.

Companies

Asked at: Cisco.

Stone Game | LeetCode 877

Medium

Array Math Dynamic Programming Game Theory

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean stoneGame(int[] piles) {
        /*
        Let's denote the number of piles as n. Since n is even, Alice will take n/2 piles and Bob will take n/2 piles.
        Let the piles be p_1, p_2, ..., p_n.
        
        Consider the sum of piles at odd indices (S_odd = p_1 + p_3 + ...) and the sum of piles at even indices (S_even = p_2 + p_4 + ...).
        (Using 1-based indexing for clarity in explanation, but code uses 0-based).
        The total number of stones is S_odd + S_even. The problem states this sum is odd.
        This means S_odd and S_even cannot be equal. One must be strictly greater than the other.

        Alice goes first. She can take either the first pile (p_1, an odd-indexed pile) or the last pile (p_n, an even-indexed pile).
        Alice has a choice between an odd-indexed pile and an even-indexed pile.

        Let's say Alice wants to secure all the odd-indexed piles. She can start by taking p_1.
        The remaining piles are p_2, ..., p_n. Now Bob must choose between p_2 and p_n, both of which are at even positions.
        No matter what Bob chooses, he will take an even-indexed pile. After his turn, the remaining piles will have an odd-indexed pile at one end and an even-indexed pile at the other.
        This means on Alice's next turn, she will again have the choice between an odd-indexed pile and an even-indexed pile.
        
        This pattern continues. Alice can always choose to take a pile from her preferred set of indices (odd or even). Her choice forces Bob to take from the other set.
        For example, if Alice decides to always take odd-indexed piles, she can. This forces Bob to always take even-indexed piles.

        Therefore, Alice can decide at the start of the game:
        1. Calculate S_odd and S_even.
        2. If S_odd > S_even, she commits to taking all the odd-indexed piles.
        3. If S_even > S_odd, she commits to taking all the even-indexed piles.

        Since S_odd != S_even, Alice can always choose the strategy that guarantees her a larger total sum of stones than Bob.
        Thus, Alice always has a winning strategy.
        */
        return true;
    }
}

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Problem Info

Description

Problem 877: Stone Game - Detailed explanation and solution approaches.

Topics

Array Math Dynamic Programming Game Theory

Companies

Cisco

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:877

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Analyze the problem constraints. The key constraint is that the total number of stones is odd, and there is an even number of piles.
2
Step 2: Realize that, given the above constraints, the sum of values at even indices and the sum of values at odd indices will never be equal.
3
Step 3: Understand that Alice always has the ability to pick either the first pile or last pile at each turn. This allows her to prioritize either even or odd indices.
4
Step 4: Conclude that Alice will always win, since she will pick the piles at either even or odd indices which give her a higher sum.
5
Step 5: Return `true`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution directly returns `true` without any iteration or data structure usage. Therefore, both time and space complexities are constant.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

A more complex approach would be to use dynamic programming to calculate the maximum difference between Alice's and Bob's scores for each subproblem. This involves creating a DP table and filling it in using a bottom-up approach.

Time: O(n^2)Space: O(n^2)

Trade-offs: While DP solves the problem, it's significantly more complex than the O(1) solution. DP is useful if the problem statement wasn't designed such that Alice always wins (e.g., there might be a cost associated with choosing specific piles). In such cases, the DP solution would allow the algorithm to determine the optimal strategy and determine whether Alice will win, and by how much.

✨

Key Insights

1Insight 1: Recognizing that the sum of stones is odd implies that the sum of stones at even indices cannot be equal to the sum of stones at odd indices.
2Insight 2: Understanding that Alice can always guarantee collecting all the stones at either the odd or even indices.
3Insight 3: The realization that Alice always wins because she can choose which set of indices to prioritize (odd or even) based on which has the larger sum, ensuring her total is greater than Bob's.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: The problem states piles.length >= 2, so it is valid for the input to have just 2 piles. The solution handles this case correctly as Alice can pick the larger of the two and win.
Edge case 2: The sum of piles[i] is odd. If this were even, the solution might not be correct, as S_odd and S_even could be equal. However, the problem statement prevents this from happening.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to simulate the game with various move combinations. This is computationally expensive and unnecessary given the game's inherent properties.

✗Common mistake 2: Overlooking the constraint that the total number of stones is odd, and the number of piles is even. These constraints are critical to understanding that Alice can always win.

✗Common mistake 3: Attempting a greedy algorithm that may not lead to the optimal solution. The strategic advantage that Alice gains by the arrangement of the stones is easily missed if a simple greedy solution is attempted.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation problems where a decision-maker can always guarantee a favorable outcome due to inherent properties of the allocation structure. This is a fairly abstract concept, but some simple games involving finite resources can mimic the Stone Game's characteristics.

🏦Real-world scenario 2: Game theory scenarios where one player has a distinct advantage based on the game's setup and rules. The advantage is often guaranteed via the constraint within the problem.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 5, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Stone Game"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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